Physics help with stopping problem

Question

You are a traffic accident investigator. You have arrived at the scene of an accident. Two trucks of equal mass(3,000 kg each) were involved in a rear-end accident at a stop sign. here is what you know: 

Truck 1 approched the intersection from the top of a 22-meter hill.

Truck 2 was on a flat stretch of road directly in front of truck 1.

At the bottom of the hill, before braking for the stop sign, Truck 1 was going 20 m/s and truck 2 was going 35 m/s.

From the skid marks on the road you can see that truck 1 applied force on its brakes for 2 seconds, 80 meters before the stop sign.

There were no skid marks left by truck 2. The collision occured at the stop sign, where truck 2 had stopped.

After the collision, both trucks were moving together in the same direction at 10 m/s, before slowly rolling to a stop.

You must now push truck 2 using 1,000 N of force, 8 meters off the side of the road

how much work was done to bring truck 2 to a stop

Answer 1

I'm not quite sure how the rest of the information plays into it, but if truck 2 was going 35 m/s and makes no attempt to stop, work needs to be done to reduce the speed of truck 2 from 35 m/s to rest at 0 m/s.

Work to stop truck 2 from speed of 35 m/s:

W = ΔKE = (.5mv²)_intial - (.5mv²)_final = .5(3000kg)(35 m/s)² - .5(3000kg)(0 m/s)² = 1,837,500 J = 1,837.5 kJ

Work to push truck 2 off the side of the road:

W = F * d = 1000N (8m) = 8,000 N*m = 8,000 J = 8 kJ



Please let me know if you have any additional questions!